Step 1: Determine the equivalent resistance for parallel resistors. The resistors $R$ and $X$ are connected in parallel. The equivalent resistance $R_{eq}$ is given by: $$R_{eq} = \frac{R \cdot X}{R + X}$$ Step 2: Apply Ohm's Law. The total voltage $E$ across the parallel combination is constant. According to Ohm's Law, $E = I \cdot R_{eq}$, where $I$ is the total current measured by the ammeter. Substituting the expression for $R_{eq}$: $$E = I \cdot \frac{R \cdot X}{R + X}$$ Step 3: Use the first set of data to form an equation for $E$. When $R = 5 \, \Omega$, the current $I = 3 \, A$. $$E = 3 \, A \cdot \frac{5 \, \Omega \cdot X}{5 \, \Omega + X}$$ $$E = \frac{15X}{5 + X} \quad (*)$$ Step 4: Use the second set of data to form another equation for $E$. When $R = 2 \, \Omega$, the current $I = 6 \, A$. $$E = 6 \, A \cdot \frac{2 \, \Omega \cdot X}{2 \, \Omega + X}$$ $$E = \frac{12X}{2 + X} \quad (**)$$ Step 5: Equate the two expressions for $E$ and solve for $X$. Since $E$ is constant, we can set equation $()$ equal to equation $(*)$: $$\frac{15X}{5 + X} = \frac{12X}{2 + X}$$ Assuming $X \neq 0$ (as $X$ is a resistance), we can divide both sides by $X$: $$\frac{15}{5 + X} = \frac{12}{2 + X}$$ Cross-multiply: $$15(2 + X) = 12(5 + X)$$ $$30 + 15X = 60 + 12X$$ Subtract $12X$ from both sides: $$30 + 15X - 12X = 60$$ $$30 + 3X = 60$$ Subtract $30$ from both sides: $$3X = 60 - 30$$ $$3X = 30$$ Divide by $3$: $$X = \frac{30}{3}$$ $$X = 10 \, \Omega$$ The value of $X$ is $10 \, \Omega$. This corresponds to option C. The final answer is $\boxed{\text{C. 10 ohms}}$.